**We want to show that given this acute triangle
that when we add the ratios on the left they are equal to 1. First
let's begin by finding the areas of certain triangles and comparing
them to the area of the whole triangle. So let's start with ratio
of the area of triangle BHC to the area of triangle ABC.**

**Now let's look at the ratio of the area
of triangle AHB to triangle ABC.**

**Finally let's look at the ratio of the area
of triangle CHB to triangle ABC.**

**What happens if we add up both sides of
the ratios of the areas??**

**Let's simplify:**

**But if we look at the triangle again, we
see that the sum of the areas of the three triangles in the numerator
is equal to the area of triangle ABC so the left hand of the equation
goes to 1 and we get the desired relation.**

**Now let's look at the second relationship.**

**First, let's look at the ratio of the area
of triangle AHB plus the area of AHC to the area of triangle ABC.**

**Next, let's look at the ratio of the area
of triangle AHC plus the area of BHC to the area of trianlge ABC.**

**Finally, let's look at the ratio of the
area of triangle AHB plus the area of BHC to the area of triangle
ABC.**

**Now if we add both sides of the ratios we
get,**

**Again note that the area of triangle AHB
plus the area of BHC plus the area of AHC is equal to the area
of ABC. So we get the desired relation.**

**Does this relation hold for an obtuse triangle?
We will use a picture to demonstrate that it doesn't.**

**If we look closely at this picture with
the orthocenter outside of triangle ABC we can see that BH is
much greater than BE and HE is much greater than BE. In fact as
angle ABC goes to 180 degrees BH and HE continue to grow larger.
This implies that we can make BH/BE > 2 by itself and HE/BE
> 1 by itself so the two statements do not hold. This concludes
the proof of the statements.**